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In the cell Zn//Zn^(-2) (c(1))//cu, E(ce...

In the cell `Zn//Zn^(-2) (c_(1))//cu, E_(cell) - E_(cell)^(0) = 0.059 V` The ratio `(C_(1))/(C_(2)) at 298 K` will be

A

`2`

B

`100`

C

`1`

D

`10^(-2)`

Text Solution

Verified by Experts

The correct Answer is:
d
`Zn + Cu^(+2) rarr Zn^(-2) + Cu`
` E_(cell)- E_(cell) = (-0.0591)/(2) log.([Zn^(+2)])/([Cu^(2+)])`brgt `0.0591= (-0.0591)/(2) log. (C_(1))/(C_(2))`
Or `log .(C_(1))/(C_(2))= "antilog" bar 2 - 10^(-2)`
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